Question: Solve for $x$ : $6\sqrt{x} - 4 = 4\sqrt{x} + 2$
Subtract $4\sqrt{x}$ from both sides: $(6\sqrt{x} - 4) - 4\sqrt{x} = (4\sqrt{x} + 2) - 4\sqrt{x}$ $2\sqrt{x} - 4 = 2$ Add $4$ to both sides: $(2\sqrt{x} - 4) + 4 = 2 + 4$ $2\sqrt{x} = 6$ Divide both sides by $2$ $\frac{2\sqrt{x}}{2} = \frac{6}{2}$ Simplify. $\sqrt{x} = 3$ Square both sides. $\sqrt{x} \cdot \sqrt{x} = 3 \cdot 3$ $x = 9$